Solutions

1. Solutions Chapter 12

2. Definitions • Solution = homogeneous mixture of two or more substances • Major component = solvent • Minor component = solute • Solubility = amount of solute that dissolves in a given volume of solvent

3. Why solutions form • Entropy: natural tendency to randomize

4. Why solutions form • Intermolecular forces

5. Like dissolves like • Polar solvents (water) dissolve polar/ionic solutes • Nonpolar solvents (oil) dissolve nonpolar solutes

6. Energy & dissolving: molecular • Separate solute (endothermic) • Separate solvent (endothermic) • Mix solute + solvent (exothermic) • DHsolution = DHsoluteDHsolvent + DHmix

7. Energy & dissolving: salts in water • Energy to separate ions (∆Hsolute) = lattice energy • ∆Hwater + ∆Hmixing = ∆Hhydration

8. Energy & dissolving: salts in water • Salts with high lattice energy may not dissolve in water • High ion charge • Small ion size • DHsolution = DHlatticeDHhydration

9. The limit of dissolving: equilibrium

10. Solubility & temperature • Most solids dissolve to higher concentrations as temperature increases • Gases are less soluble as temperature increases

11. Gas solubility & pressure • Gas solubility increases as the pressure of that gas increases • Henry’s Law

12. Tro, Chemistry: A Molecular Approach Henry’s Law • the solubility of a gas (Sgas) is directly proportional to its partial pressure, (Pgas) Sgas = kHPgas • kH is called Henry’s Law Constant

14. Tro, Chemistry: A Molecular Approach Concentrations • solutions have variable composition • to describe a solution, need to describe components and relative amounts • the terms dilute and concentrated can be used as qualitative descriptions of the amount of solute in solution • concentration = amount of solute in a given amount of solution • occasionally amount of solvent

15. Tro, Chemistry: A Molecular Approach moles of solute liters of solution molarity = Solution ConcentrationMolarity • moles of solute per 1 liter of solution • used because it describes how many molecules of solute in each liter of solution • if a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar

16. Tro, Chemistry: A Molecular Approach Solution ConcentrationMolality, m • moles of solute per 1 kilogram of solvent • defined in terms of amount of solvent, not solution • like the others • does not vary with temperature • because based on masses, not volumes

17. Tro, Chemistry: A Molecular Approach Percent • parts of solute in every 100 parts solution • mass percent = mass of solute in 100 parts solution by mass • if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution • or 0.9 kg solute in every 100 kg solution

18. Tro, Chemistry: A Molecular Approach Percent Concentration

19. Tro, Chemistry: A Molecular Approach Using Concentrations asConversion Factors • concentrations show the relationship between the amount of solute and the amount of solvent • 12%(m/m) sugar(aq) means 12 g sugar  100 g solution • or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution • 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution • 22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution • The concentration can then be used to convert the amount of solute into the amount of solution, or vice versa

20. Tro, Chemistry: A Molecular Approach Preparing a Solution • need to know amount of solution and concentration of solution • calculate the mass of solute needed • start with amount of solution • use concentration as a conversion factor • 5% by mass Þ 5 g solute  100 g solution • “Dissolve the grams of solute in enough solvent to total the total amount of solution.”

21. grams solute grams solution x 106 Solution ConcentrationPPM • grams of solute per 1,000,000 g of solution • mg of solute per 1 kg of solution • 1 liter of water = 1 kg of water • for water solutions we often approximate the kg of the solution as the kg or L of water mg solute kg solution mg solute L solution

22. Tro, Chemistry: A Molecular Approach moles of components A total moles in the solution mole fraction of A = XA = Solution ConcentrationsMole Fraction, XA • the mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution • total of all the mole fractions in a solution = 1 • unitless • the mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution • = mole fraction x 100%

23. Tro, Chemistry: A Molecular Approach Raoult’s Law • the vapor pressure of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P° Psolvent in solution = csolvent∙P° • since the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent

24. Tro, Chemistry: A Molecular Approach Ionic Solutes and Vapor Pressure • according to Raoult’s Law, the effect of solute on the vapor pressure simply depends on the number of solute particles • when ionic compounds dissolve in water, they dissociate – so the number of solute particles is a multiple of the number of moles of formula units • the effect of ionic compounds on the vapor pressure of water is magnified by the dissociation • since NaCl dissociates into 2 ions, Na+ and Cl, one mole of NaCl lowers the vapor pressure of water twice as much as 1 mole of C12H22O11 molecules would

25. Tro, Chemistry: A Molecular Approach Effect of Dissociation

26. Tro, Chemistry: A Molecular Approach cH2O, P°H2O PH2O Ex 12.6 – What is the vapor pressure of H2O when 0.102 mol Ca(NO3)2 is mixed with 0.927 mol H2O @ 55°C? Given: Find: 0.102 mol Ca(NO3)2, 0.927 mol H2O, P° = 118.1 torr P of H2O, torr Concept Plan: Relationships: P = c∙P° Solve: Ca(NO3)2 Ca2+ + 2 NO3  3(0.102 mol solute) Check: the unit is correct, the pressure lower than the normal vapor pressure makes sense

27. Tro, Chemistry: A Molecular Approach Raoult’s Law for Volatile Solute • when both the solvent and the solute can evaporate, both molecules will be found in the vapor phase • the total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent • for an ideal solution Ptotal = Psolute + Psolvent • the solvent decreases the solute vapor pressure in the same way the solute decreased the solvent’s Psolute = csolute∙P°solute and Psolvent = csolvent∙P°solvent

28. Tro, Chemistry: A Molecular Approach Ideal vs. Nonideal Solution • in ideal solutions, the made solute-solvent interactions are equal to the sum of the broken solute-solute and solvent-solvent interactions • ideal solutions follow Raoult’s Law • effectively, the solute is diluting the solvent • if the solute-solvent interactions are stronger or weaker than the broken interactions the solution is nonideal

29. Tro, Chemistry: A Molecular Approach Vapor Pressure of a Nonideal Solution • when the solute-solvent interactions are stronger than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be less than predicted by Raoult’s Law • because the vapor pressures of the solute and solvent are lower than ideal • when the solute-solvent interactions are weaker than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be larger than predicted by Raoult’s Law

30. Tro, Chemistry: A Molecular Approach Deviations from Raoult’s Law

31. Tro, Chemistry: A Molecular Approach

32. Tro, Chemistry: A Molecular Approach Freezing Point Depression • the freezing point of a solution is lower than the freezing point of the pure solvent • for a nonvolatile solute • therefore the melting point of the solid solution is lower • the difference between the freezing point of the solution and freezing point of the pure solvent is directly proportional to the molal concentration of solute particles (FPsolvent – FPsolution) = DTf = m∙Kf • the proportionality constant is called the Freezing Point Depression Constant, Kf • the value of Kf depends on the solvent • the units of Kf are °C/m

33. Tro, Chemistry: A Molecular Approach Kf

34. Tro, Chemistry: A Molecular Approach m DTf Ex 12.8 – What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C2H6O2? Given: Find: 1.7 m C2H6O2(aq) Tf, °C Concept Plan: Relationships: DTf = m ∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00°C Solve: Check: the unit is correct, the freezing point lower than the normal freezing point makes sense

35. Tro, Chemistry: A Molecular Approach Boiling Point Elevation • the boiling point of a solution is higher than the boiling point of the pure solvent • for a nonvolatile solute • the difference between the boiling point of the solution and boiling point of the pure solvent is directly proportional to the molal concentration of solute particles (BPsolution – BPsolvent) = DTb = m∙Kb • the proportionality constant is called the Boiling Point Elevation Constant, Kb • the value of Kb depends on the solvent • the units of Kb are °C/m

36. Tro, Chemistry: A Molecular Approach DTb m kg H2O mol C2H6O2 g C2H6O2 Ex 12.9 – How many g of ethylene glycol, C2H6O2, must be added to 1.0 kg H2O to give a solution that boils at 105°C? Given: Find: 1.0 kg H2O, Tb = 105°C mass C2H6O2, g Concept Plan: Relationships: DTb = m ∙Kb, Kb H2O = 0.512 °C/m, BPH2O = 100.0°C MMC2H6O2 = 62.07 g/mol, 1 kg = 1000 g Solve:

37. Tro, Chemistry: A Molecular Approach Osmosis • osmosis is the flow of solvent through a semi-permeable membrane from solution of low concentration to solution of high concentration • the amount of pressure needed to keep osmotic flow from taking place is called the osmotic pressure • the osmotic pressure, P, is directly proportional to the molarity of the solute particles • R = 0.08206 (atm∙L)/(mol∙K) P = MRT

38. Tro, Chemistry: A Molecular Approach

39. Tro, Chemistry: A Molecular Approach P,T M mL L mol protein Ex 12.10 – What is the molar mass of a protein if 5.87 mg per 10 mL gives an osmotic pressure of 2.45 torr at 25°C? Given: Find: 5.87 mg/10 mL, P = 2.45 torr, T = 25°C molar mass, g/mol Concept Plan: Relationships: P=MRT, T(K)=T(°C)+273.15, R= 0.08206atm∙L/mol∙K M = mol/L, 1 mL = 0.001 L, MM = g/mol, 1 atm = 760 torr Solve:

40. Tro, Chemistry: A Molecular Approach Colligative Properties • colligative properties are properties whose value depends only on the number of solute particles, and not on what they are • Vapor Pressure Depression, Freezing Point Depression, Boiling Point Elevation, Osmotic Pressure • the van’t Hoff factor, i, is the ratio of moles of solute particles to moles of formula units dissolved • measured van’t Hoff factors are often lower than you might expect due to ion pairing in solution

41. Tro, Chemistry: A Molecular Approach

42. Tro, Chemistry: A Molecular Approach A hyperosmotic solution has a higher osmotic pressure than the solution inside the cell – as a result there is a net flow of water out of the cell, causing it to shrivel A hyposmotic solution has a lower osmotic pressure than the solution inside the cell – as a result there is a net flow of water into the cell, causing it to swell An isosmotic solution has the same osmotic pressure as the solution inside the cell – as a result there is no net the flow of water into or out of the cell.

43. Tro, Chemistry: A Molecular Approach Colloids • a colloidal suspension is a heterogeneous mixture in which one substance is dispersed through another • most colloids are made of finely divided particles suspended in a medium • the difference between colloids and regular suspensions is generally particle size – colloidal particles are from 1 to 100 nm in size

44. Tro, Chemistry: A Molecular Approach Properties of Colloids • the particles in a colloid exhibit Brownian motion • colloids exhibit the Tyndall Effect • scattering of light as it passes through a suspension • colloids scatter short wavelength (blue) light more effectively than long wavelength (red) light

45. Tro, Chemistry: A Molecular Approach

46. Tro, Chemistry: A Molecular Approach

47. Tro, Chemistry: A Molecular Approach Soap and Micelles • soap molecules have both a hydrophilic (polar/ionic) “head” and a hydrophobic (nonpolar) “tail” • part of the molecule is attracted to water, but the majority of it isn’t • the nonpolar tail tends to coagulate together to form a spherical structure called a micelle

48. Tro, Chemistry: A Molecular Approach

49. Tro, Chemistry: A Molecular Approach Soap Action • most dirt and grease is made of nonpolar molecules – making it hard for water to remove it from the surface • soap molecules form micelles around the small oil particles with the polar/ionic heads pointing out • this allows the micelle to be attracted to water and stay suspended

50. Tro, Chemistry: A Molecular Approach The polar heads of the micelles attract them to the water, and simultaneously repel other micelles so they will not coalesce and settle out.